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postnew [5]
4 years ago
8

Helppppppppppppppppppppppppppppppppppppp

Mathematics
2 answers:
Fed [463]4 years ago
8 0

Answer:its 8

Step-by-step explanation:

egoroff_w [7]4 years ago
5 0

Step-by-step explanation:

rng

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My number has 2hundreds. The tens digit is 9more than the ones digit. My number is=
levacccp [35]
290 
There's 2 hundreds = 200
Then you have 9 in the tens digit.
The only possible answer for the one's digit is 0 since 0 + 9 = 9.
So the answer is 290. 
5 0
3 years ago
40 POINTS answer correctly
Alexxx [7]
Since the denominator is the number on the bottom, we need to find a common denominator for a and 5a^2. Since a*5a is 5a^2, we have a common denominator - 5a^2! For the fractions, that means that since a*5a=5a^2, we have to multiply -7 by 5a to get -35a/5a^2-9/5a^2=(-35a-9)/5a^2
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3 years ago
Read 2 more answers
What will be the new coordinate of vertex B if the triangle is dilated with a center at the origin by a scale factor of 1/3
a_sh-v [17]

You must observe the distances from the center of the dilation at point A to the other points B, C and D. The dilation image will be 1/3 of each of these distances. AB = 6, so A'B' = 2. AD = 9, so A'D' = 3.

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7 0
3 years ago
What is the discriminant of the quadratic equation 0= -x^2+4x-2
xxTIMURxx [149]
The discrimination is 8
3 0
4 years ago
A construction firm bids on two different contracts. Let E1 be the event that the bid on the first contract is successful, and d
NARA [144]

Answer:

(a) 0.56

(b) 0.06

(c) 0.94

Step-by-step explanation:

P(E1) = 0.7 and P(E2) = 0.8

(a) The probability that both bids are successful is given by the product of the probability of success of each bid:

P(E1\ and\ E2) = 0.7*0.8=0.56

(b) The probability that neither bid is successful is given by the product of the probability of failure of each bid:

P(not\ E1\ and\ not\ E2)= (1-P(E1))*(1-P(E2))\\P(not\ E1\ and\ not\ E2)=0.3*0.2=0.06

(c) The probability that the firm is successful in at least one of the two bids is given by the sum of the probability of success of each bid subtracted by the probability that both bids are successful:

P(E1\ or\ E2)=P(E1)+P(E2) - P(E1\ and\ E2)\\P(E1\ or\ E2)=0.7+0.8-0.56\\P(E1\ or\ E2)=0.94

3 0
3 years ago
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